The E-Cat as a Cavity Resonator Producing 50-120keV Radiation With Side Effects (Paul Dodgshun)

The following post has been submitted by Paul Dodgshun

The E-cat as a cavity resonator producing 50-120keV radiation with side effects.

Define machine :

‘A machine is a tool containing one or more parts that uses energy to perform an intended action.’
[following the definition, a helpful sentence]

‘Machines are usually powered by mechanical, chemical, THERMAL, or ELECTRICAL means, …’

The essence of a machine is to use energy (in all its forms) for a defined purpose.

Part 2 – The Theory
The main effect is the thermalization of low-energy Gamma (50-120 keV).
The transmutation of Nickel into Copper is a side effect.

The cavity magnetron outputs microwaves using the interaction of a stream of electrons with a magnetic field while moving past a series of open metal cavities (cavity resonators)

So you need an oscillator that produces ‘low-energy Gamma (50-120 keV)’.

What oscillator would do that?

Free Electron Laser Oscillator (FELO) – An FELO for hard X-Rays

A low-gain device with a low-loss x-ray cavity

Where have we got cavities in the Nickel that could oscillate? We have just been discussing that. The cavities in the Nickel generated by very high pressure H2 and close to the Nickel surface where EMF stimulation will work. The cavities absorb some of their own heat output to keep the Nickel and Hydrogen atoms/molecules vibrating as the cavities emit energy as radiation.

A resonant cavity is a MACHINE. It uses heat and EMF stimulation to output 50-120keV radiation. Energy inputs and outputs defined. In the case of the E-cat, the 50-120keV radiation is dumped in the lead shielding, the lead heats the water and boils it into steam. That is CoP=6 sorted.

Next figure what kicks off p+Li7->2He4+17.3MeV. From Piantelli :-
1: Orbital Capture
(H-)-2e -> p

2: Nuclear Capture
p+Ni -> Cu
-1H + 58Ni-> 59Cu + 3.417 MeV{1a}
-1H + 60Ni-> 61Cu + 4.796 MeV{1b}
-1H + 61Ni-> 62Cu + 5.866 MeV{1c}
-1H + 62Ni-> 63Cu + 6.122 MeV{1d}
-1H + 64Ni-> 65Cu + 7.453 MeV{1e}

3: Coulomb Repulsion
p+Ni -> p+Ni 6.7Mev

Repelled ‘p’s
3.1 p+Ni -> Cu Nuclear Capture as in 2:
3.2 p+Li7 -> 2He4 Lithium Fission

Why cannot (H-)-2e -> p feed directly into p+Li7 -> 2He4 as well as p+Ni -> Cu?

He4+8.65MeV hits H2 and produces one of H0+H2 or two of H1 (where 0,1,2 is the number of attached electrons).
He4 is also looking for electrons to form a neutral atom. H0 is a proton p and the H1 a heavy electron that might reach the Li nucleus through an electron cloud of just seven electrons.

That is a possible four protons out for two He collisions with two H2 molecules from just one p+Li7 event.
That has the makings of a chain reaction !! Now figure out the control scheme to get the number of ‘p’s in one
generation the same as in the next, so the reactor can run in a stable fashion.